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Question Number 193436 by beto last updated on 14/Jun/23

$$$$$$\frac{sin(x+{18}^o)}{sin\left({48}^o\right)}=\frac{sin\left(x\right)}{sin\left({18}^o\right)}$$$$$$

Answered by a.lgnaoui last updated on 14/Jun/23

$$\sin (\mathrm{x}+18)=\frac{\sin 48}{\sin 18}\sin \mathrm{x}$$$$\sin \mathrm{xcos}18+\cos \mathrm{xsin}18=\frac{\sin 48}{\sin 18}\sin \mathrm{x}$$$$\cos \mathrm{x}[\tan \mathrm{xcos}18+\sin 18]=\cos \mathrm{x}\left[\frac{\sin 48}{\sin 18}\tan \mathrm{x}\right]$$$$\Rightarrow \tan \mathrm{xcos}18+\sin 18=\frac{\sin 48}{\sin 18}\tan \mathrm{x}$$$$\tan \mathrm{x}(\frac{\sin 48}{\sin 18}-\cos 18)=\sin 18$$$$\tan \mathrm{x}(\sin 48-\sin 18\cos 18)=\sin {}^{2}18$$$$\Rightarrow \tan \mathbf{x}=\frac{{\sin }^{2}18}{\sin 48-\sin 18\cos 18}$$$$=0,21255656167$$$$$$$$\mathbf{x}=12$$$$$$

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